Trigonometry

Find the right-angle square. The hypotenuse is the side facing that right angle.

Identify the two labelled sides relative to the 35-degree angle, then choose the ratio that links them.

Choose the ratio linking the marked side and the hypotenuse. Use degree mode.

Identify which labelled side touches the given angle and is not the hypotenuse.

The labelled sides meet at the right angle. Which trig ratio links them?

Write the ratio equation first. Here the unknown is underneath the fraction.

1. The 8.4 cm side is opposite 52° and x is adjacent.
2. \tan52°=\dfrac{8.4}{x}.
3. Multiply by x and divide by \tan52°: x=\dfrac{8.4}{\tan52°}.
4. x=6.562799263\ldots cm, so x=6.6 cm.

The unknown is the longest side. Set up the ratio before rearranging.

1. The side 6.8 cm is opposite 28°, and x is the hypotenuse.
2. \sin28°=\dfrac{6.8}{x}, so x=\dfrac{6.8}{\sin28°}.
3. x=14.48437038\ldots cm, giving 14.5 cm.

Link the known side touching the angle to the unknown hypotenuse.

1. The 11 cm side is adjacent to 35°, and x is the hypotenuse.
2. \cos35°=\dfrac{11}{x}, so x=\dfrac{11}{\cos35°}.
3. x=13.42852048\ldots cm, so x=13.4 cm.

Find a ratio from the two given lengths, then use an inverse trig function in degree mode.

1. The 7 cm side is opposite x, and the hypotenuse is 12 cm.
2. \sin x=\dfrac{7}{12}.
3. x=\sin^{-1}\left(\dfrac{7}{12}\right)=35.68533471\ldots°.
4. To 1 decimal place, x=35.7°.

The given lengths are the adjacent side and the hypotenuse. Use the corresponding inverse function.

1. The 9 cm side is adjacent to x, and 15 cm is the hypotenuse.
2. \cos x=\dfrac{9}{15}.
3. x=\cos^{-1}\left(\dfrac{9}{15}\right)=53.13010235\ldots°.
4. So x=53.1°.

Neither given side is the hypotenuse. Compare their positions relative to x.

1. The 6.3 cm side is opposite x, and 9.2 cm is adjacent.
2. \tan x=\dfrac{6.3}{9.2}.
3. x=\tan^{-1}\left(\dfrac{6.3}{9.2}\right)=34.40266485\ldots°.
4. So x=34.4°.

The ladder is the hypotenuse. Identify the side representing the height.

1. The wall and level ground form a right angle. The height h is opposite 68°.
2. \sin68°=\dfrac{h}{4.8}, so h=4.8\sin68°.
3. h=4.450482502\ldots m, which is 4.45 m to 3 significant figures.

Use the vertical height and horizontal distance to find the angle at the anchor.

1. The pole and ground form a right angle. Relative to the ground angle x, the height 5.6 m is opposite and 3.4 m is adjacent.
2. \tan x=\dfrac{5.6}{3.4}.
3. x=\tan^{-1}\left(\dfrac{5.6}{3.4}\right)=58.73626831\ldots°.
4. The angle is 59° to the nearest degree.

Use the right triangle formed by the diagonal. Identify the side facing the marked angle.

1. The side opposite x is 9 cm, and the diagonal is the hypotenuse of length 15 cm.
2. \sin x=\dfrac{9}{15}.
3. x=\sin^{-1}\left(\dfrac{9}{15}\right)=36.86989765\ldots°.
4. So x=36.9°.

Calculate the horizontal length in metres first, then convert it to centimetres.

1. The rise is opposite 12° and the horizontal length is adjacent.
2. Let d be the horizontal length in metres. Then \tan12°=\dfrac{0.9}{d}, so d=\dfrac{0.9}{\tan12°}=4.234167099\ldots m.
3. Convert to centimetres: x=100d=423.4167099\ldots cm.
4. The horizontal length is 423 cm to the nearest centimetre.

Draw a perpendicular from the top vertex to the base in your working. It splits the triangle into two equal right-angled triangles.

1. Draw a perpendicular from the top vertex to the base. In an isosceles triangle this also halves the base.
2. The height is h=12\sin42°=8.029567276\ldots cm.
3. Half the base is u=12\cos42°=8.917737906\ldots cm, so the full base is 2u.
4. Area =\dfrac12\times(2u)\times h=u h.
5. Using the unrounded values, area =71.60557647\ldots\text{ cm}^2, which is 71.6\text{ cm}^2 to 3 significant figures.

In the right-angled triangle BCD, BD is opposite the 27° angle and BC is next to it. Find BC first, then add AB.

1. In triangle BCD, BD =6 m is opposite angle C and BC is adjacent to it, so use tan.
2. \tan27°=\dfrac{6}{BC}, so BC=\dfrac{6}{\tan27°}=11.7756\ldots m.
3. Since B lies between A and C, AC=AB+BC=8+11.7756\ldots=19.7756\ldots m.
4. To 1 decimal place, AC=19.8 m.

Triangle ABP has a right angle at B. Use it to find BP, then remember that BC is 9 cm.

1. Triangle ABP has a right angle at B. BP is opposite the 25° angle and AB =12 cm is adjacent to it, so use tan.
2. \tan25°=\dfrac{BP}{12}, so BP=12\tan25°=5.5957\ldots cm.
3. PC=BC-BP=9-5.5957\ldots=3.4043\ldots cm.
4. To 1 decimal place, PC=3.4 cm.

Think about which allowed angle gives the shortest horizontal run. Compare using the unrounded result.

Part (a)

1. For a fixed height, making the angle smaller needs a longer horizontal run. The smallest run therefore uses the largest allowed angle, 8°.
2. \tan8°=\dfrac{1.4}{d}, so d=\dfrac{1.4}{\tan8°}=9.961517611\ldots m.
3. This is 10.0 m to 1 decimal place.

Part (b)

1. The minimum required run is 9.961517611\ldots m, which is more than the available 9.8 m.
2. So the ramp cannot fit while meeting the angle requirement.

Find both unlabelled sides, using the same given angle. Add all three side lengths without rounding early.

1. The side 8.4 cm is opposite 32°.
2. For the hypotenuse h, \sin32°=\dfrac{8.4}{h}, so h=\dfrac{8.4}{\sin32°}=15.85147128\ldots cm.
3. For the other short side a, \tan32°=\dfrac{8.4}{a}, so a=\dfrac{8.4}{\tan32°}=13.44281004\ldots cm.
4. Using unrounded lengths, perimeter =8.4+h+a=37.69428133\ldots cm.
5. To 3 significant figures, the perimeter is 37.7 cm.